Question

In Fig. 35-4, assume that the two light waves, of wavelength $\mathbf{620}\mathbf{}\mathbf{\text{nm}}$ in air, are initially out of phase by $\mathbf{\pi }$ rad. The indexes of refraction of the media are ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}$ and${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{65}$ . What are the (a) smallest and (b) second smallest value of $\mathit{L}$that will put the waves exactly in phase once they pass through the two media?

Short answer

(a) The smallest value of $L$is $1.55\times {10}^{-6}\text{m}$ .

(b) The second smallest value of L is $4.65\times {10}^{-6}\text{m}$.

Step-by-step solution

Write the given data from the question:

The wavelength in air,$\lambda =620\text{nm}$

The phase difference,$\Delta \varphi =\pi$

The refractive index of medium 1, $n_1=1.45$

The refractive index of medium 2, $n_2=1.65$

Determine the formulas to calculate the smallest and second smallest value of L:

The expression to calculate the phase difference is given as follows.

$\mathit{\Delta }\mathit{\varphi }\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{L}\left(\frac{n_2}{\lambda }-\frac{n_1}{\lambda }\right)$

$\mathit{L}\mathbf{=}\frac{\mathbf{\lambda }\mathbf{\Delta }\mathbf{\varphi }}{\mathbf{2}\mathbf{\pi }\left(n_2-n_1\right)}$ ......(1)

The expression to calculate the second smallest value of Lis given as follows.

$\mathit{L}\mathbf{=}\frac{\mathbf{3}\mathbf{\lambda }\mathbf{\Delta }\mathbf{\varphi }}{\mathbf{2}\mathbf{\pi }\left(n_2-n_1\right)}$

(a) Calculate the smallest value of L  :

Calculate the smallest value of the L .

Substitute the given values into equation (1).

$\begin{array}{rcl}L& =& \frac{\pi \times 620\times {10}^{-9}}{2\pi \left(1.65-1.45\right)}\\ & =& \frac{310\times {10}^{-9}}{0.20}\\ & =& 1550\times {10}^{-9}\\ & =& 1.55\times {10}^{-6}\text{m}\end{array}$

Hence, the smallest value of $L$ is $1.55\times {10}^{-6}\text{m}$

(b) Calculate the second smallest value of L  :

Calculate the smallest value of the$L$.

Substitute the given values into equation (1).

$\begin{array}{rcl}L& =& \frac{3\pi \times 620\times {10}^{-9}}{2\pi \left(1.65-1.45\right)}\\ & =& \frac{930\times {10}^{-9}}{0.20}\\ & =& 4650\times {10}^{-9}\\ & =& 4.65\times {10}^{-6}\text{m}\end{array}$

Hence the second smallest value of $L$ is $4.65\times {10}^{-6}\text{m}$.